I am trying perform a function to each cell of a matrix in R. I want to add the cells per 3 if they are > 0.
Example:
mat <- matrix(data=0:9, nrow=5, ncol=10, byrow=F)
mat3 <- apply(mat, MARGIN = 1, FUN= function(mat) if(mat != 0) {mat+3})
But first that created a list of length 5 and second it's all the cells who are added per 3.
For this simple case, it would be preferable to use the solutions from #akrun or #Karolis Koncevičius, but you can also do:
apply(mat, 2, function(x) ifelse(x > 0, x + 3, x))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 8 0 8 0 8 0 8 0 8
[2,] 4 9 4 9 4 9 4 9 4 9
[3,] 5 10 5 10 5 10 5 10 5 10
[4,] 6 11 6 11 6 11 6 11 6 11
[5,] 7 12 7 12 7 12 7 12 7 12
You don't need any apply, can use ifelse directly:
ifelse(mat > 0, mat+3, mat)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 8 0 8 0 8 0 8 0 8
[2,] 4 9 4 9 4 9 4 9 4 9
[3,] 5 10 5 10 5 10 5 10 5 10
[4,] 6 11 6 11 6 11 6 11 6 11
[5,] 7 12 7 12 7 12 7 12 7 12
But a faster solution would be:
mat[mat > 0] <- mat[mat > 0] + 3
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 8 0 8 0 8 0 8 0 8
[2,] 4 9 4 9 4 9 4 9 4 9
[3,] 5 10 5 10 5 10 5 10 5 10
[4,] 6 11 6 11 6 11 6 11 6 11
[5,] 7 12 7 12 7 12 7 12 7 12
We could also do this on the fly with
mat + (mat > 0) * 3
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#[1,] 0 8 0 8 0 8 0 8 0 8
#[2,] 4 9 4 9 4 9 4 9 4 9
#[3,] 5 10 5 10 5 10 5 10 5 10
#[4,] 6 11 6 11 6 11 6 11 6 11
#[5,] 7 12 7 12 7 12 7 12 7 12
Related
I was able to write a function in r to "shift" a column of a matrix over to the right by one:
shift <- function(disc){
mat <- matrix(nrow = 4, ncol = 12)
mat[,1] <- disc[,12]
for(i in 1:11){
mat[,i+1] <- disc[,i]
}
return(mat)
}
So to see how that works:
> disc0
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 2 5 10 7 16 8 7 8 8 3 4 12
[2,] 3 3 14 14 21 21 9 9 4 4 6 6
[3,] 8 9 10 11 12 13 14 15 4 5 6 7
[4,] 14 11 14 14 11 14 11 14 11 11 14 11
> shift(disc0)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 12 2 5 10 7 16 8 7 8 8 3 4
[2,] 6 3 3 14 14 21 21 9 9 4 4 6
[3,] 7 8 9 10 11 12 13 14 15 4 5 6
[4,] 11 14 11 14 14 11 14 11 14 11 11 14
What if I wanted to shift over 3 times, for example? I could do this manually:
> x <- disc0
> x <- shift(x)
> x <- shift(x)
> x <- shift(x)
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 3 4 12 2 5 10 7 16 8 7 8 8
[2,] 4 6 6 3 3 14 14 21 21 9 9 4
[3,] 5 6 7 8 9 10 11 12 13 14 15 4
[4,] 11 14 11 14 11 14 14 11 14 11 14 11
So now the original first column (2,3,8,14) is now in the 4th column.
But how can I automate this? I want to write a function that will repeat my shift function n times. Thanks in advance
You could write a function that takes in the shift parameter:
shift <- function(x, num = 1){
n <- ncol(x)
x[, c((n - num +1):n, 1:(n - num))]
}
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 2 3 4 5 6 7 8
[2,] 1 2 3 4 5 6 7 8
[3,] 1 2 3 4 5 6 7 8
[4,] 1 2 3 4 5 6 7 8
shift(mat)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 8 1 2 3 4 5 6 7
[2,] 8 1 2 3 4 5 6 7
[3,] 8 1 2 3 4 5 6 7
[4,] 8 1 2 3 4 5 6 7
shift(mat,2)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 7 8 1 2 3 4 5 6
[2,] 7 8 1 2 3 4 5 6
[3,] 7 8 1 2 3 4 5 6
[4,] 7 8 1 2 3 4 5 6
shift(mat,3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 6 7 8 1 2 3 4 5
[2,] 6 7 8 1 2 3 4 5
[3,] 6 7 8 1 2 3 4 5
[4,] 6 7 8 1 2 3 4 5
You may use a for loop -
n <- 3
for(i in seq_len(n)) {
disc0 <- shift(disc0)
}
disc0
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,] 3 4 12 2 5 10 7 16 8 7 8 8
#[2,] 4 6 6 3 3 14 14 21 21 9 9 4
#[3,] 5 6 7 8 9 10 11 12 13 14 15 4
#[4,] 11 14 11 14 11 14 14 11 14 11 14 11
I have an array of number
x <- seq(1:10)
I am after a matrix with n rows. Here is an example with 3-row matrix:
1 2 3 4 5 6 7 8 9 10
NA 1 2 3 4 5 6 7 8 9
NA NA 1 2 3 4 5 6 7 8
What would be the best way to create one?
There is an odd little function called embed that will do it...
t(embed(c(NA, NA, 1:10), 3))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 2 3 4 5 6 7 8 9 10
[2,] NA 1 2 3 4 5 6 7 8 9
[3,] NA NA 1 2 3 4 5 6 7 8
For a vector x and a matrix of n rows, the equivalent would be
t(embed(c(rep(NA, n-1), x), n))
Maybe there is more simpler way to do this but one way to create this matrix would be
create_matrix <- function(x, n) {
t(sapply(seq(n), function(m) c(rep(NA, m - 1), head(x, length(x) - m + 1))))
}
create_matrix(1:10, 3)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#[1,] 1 2 3 4 5 6 7 8 9 10
#[2,] NA 1 2 3 4 5 6 7 8 9
#[3,] NA NA 1 2 3 4 5 6 7 8
create_matrix(c(4, 3, 6, 8, 7), 4)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 4 3 6 8 7
#[2,] NA 4 3 6 8
#[3,] NA NA 4 3 6
#[4,] NA NA NA 4 3
I'm doing cross validation and I want to separate the data into 3 folds.
I create a matrix withmat=matrix(sample.int(10, 9*100, TRUE), 6, 10) which looks like this:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 10 10 9 3 3 3 4 4 3 9
[2,] 9 3 5 1 3 9 5 5 4 8
[3,] 7 6 6 3 8 2 3 10 7 4
[4,] 7 4 10 8 7 5 2 6 2 8
[5,] 9 7 7 5 3 9 5 8 7 8
[6,] 3 3 1 2 9 3 6 7 6 9
I want to get then 3 matrices with the data:
fold 1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 10 10 9 3 3 3 4 4 3 9
[2,] 9 3 5 1 3 9 5 5 4 8
fold 2
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 7 6 6 3 8 2 3 10 7 4
[2,] 7 4 10 8 7 5 2 6 2 8
fold 3
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 9 7 7 5 3 9 5 8 7 8
[2,] 3 3 1 2 9 3 6 7 6 9
Here is my code what I did:
require(stats)
mat=matrix(sample.int(10, 9*100, TRUE), 6, 10)
folds=cut(seq(1, nrow(mat)), breaks = 3, labels = FALSE)
#Perform 10 fold cross validation
for(i in 1:3){
#segment your data by folds using the which() function
testIndexes=which(folds==i, arr.ind = TRUE)
testData=mat[testIndexes,]
trainData=mat[-testIndexes,]
}
The training data that I get from fold 1 and fold 2 are connected, I want to generate them separately.
This is the generated training set which should be separate in two folds.
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 10 10 9 3 3 3 4 4 3 9
[2,] 9 3 5 1 3 9 5 5 4 8
[3,] 7 6 6 3 8 2 3 10 7 4
[4,] 7 4 10 8 7 5 2 6 2 8
Each loop of my sapply function will out put a n*m matrix. n is fixed, m is not.
For example, if I run this in R:
sapply(1:3, function(x) {matrix(1:9, 3)})
and it will output:
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 4 4
[5,] 5 5 5
[6,] 6 6 6
[7,] 7 7 7
[8,] 8 8 8
[9,] 9 9 9
However, what I want is something like this:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 1 4 7 1 4 7 1 4 7
[2,] 2 5 8 2 5 8 2 5 8
[3,] 3 6 9 3 6 9 3 6 9
Any idea for this? Thanks
One solution is:
do.call(cbind, lapply(1:3, function(x) {matrix(1:9, 3)}))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 1 4 7 1 4 7 1 4 7
[2,] 2 5 8 2 5 8 2 5 8
[3,] 3 6 9 3 6 9 3 6 9
We can use replicate
`dim<-`(replicate(3, matrix(1:9, 3)), c(3, 3*3))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
#[1,] 1 4 7 1 4 7 1 4 7
#[2,] 2 5 8 2 5 8 2 5 8
#[3,] 3 6 9 3 6 9 3 6 9
I found this while searching for a similar approach.
Selecting rows with same result in different columns in R
Is there a way to search within a range of columns? Playing off the example in the link, what if instead of catch[catch$tspp.name == catch$elasmo.name,], is it possible to do this?
catch[catch$tspp.name == c[23:56],] where R would search for values within columns 23 to 56 that match the tspp value?
Thanks in advance and please let me know whether it's better to post an independent question on a topic related to a previous post or to insert a follow up question within the aforementioned post.
Here's one way to do it. This finds rows of X where the first column appears in columns 2 through 9.
> set.seed(1)
> X<-matrix(sample(10,100,T),10)
> X
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 3 3 10 5 9 5 10 4 5 3
[2,] 4 2 3 6 7 9 3 9 8 1
[3,] 6 7 7 5 8 5 5 4 4 7
[4,] 10 4 2 2 6 3 4 4 4 9
[5,] 3 8 3 9 6 1 7 5 8 8
[6,] 9 5 4 7 8 1 3 9 3 8
[7,] 10 8 1 8 1 4 5 9 8 5
[8,] 7 10 4 2 5 6 8 4 2 5
[9,] 7 4 9 8 8 7 1 8 3 9
[10,] 1 8 4 5 7 5 9 10 2 7
> X[rowSums(X[,1]==X[,2:9])>0,]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 3 3 10 5 9 5 10 4 5 3
[2,] 3 8 3 9 6 1 7 5 8 8
[3,] 9 5 4 7 8 1 3 9 3 8
[4,] 7 4 9 8 8 7 1 8 3 9